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Intermediate Geometry : Intermediate Geometry

Study concepts, example questions & explanations for Intermediate Geometry

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All Intermediate Geometry Resources

8 Diagnostic Tests 250 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #6 : How To Find The Area Of An Acute / Obtuse Isosceles Triangle

An isosceles triangle is placed in a circle as shown by the figure below.

If the diameter of the circle is , find the area of the shaded region.

Possible Answers:

190.06

328.52

222.49

214.16

Correct answer:

214.16

Explanation:

From the given image, you should notice that the base of the triangle is also the diameter of the circle. In addition, the height of the triangle is also the radius of the circle.

Thus, we can find the area of the triangle.

$\text{Area of Triangle}=\frac{1}{2}(\text{base}\times\text{height})$

$\text{Area of Triangle}=\frac{1}{2}(20 \times 10)=100$

Next, recall how to find the area of a circle.

$\text{Area of Circle}=\pi\times\text{radius}^2$

$\text{Area of Circle}=\pi\times 10^2=100 \pi$

To find the area of the shaded region, subtract the two areas.

$\text{Area of Shaded Region}=\text{Area of Circle}-\text{Area of Triangle}$

$\text{Area of Shaded Region}=100\pi-100=214.16$

Make sure to round to places after the decimal.

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Example Question #7 : How To Find The Area Of An Acute / Obtuse Isosceles Triangle

An isosceles triangle is placed in a circle as shown by the figure below.

If the diameter of the circle is , find the area of the shaded region.

Possible Answers:

420.94

458.61

398.50

419.75

Correct answer:

419.75

Explanation:

From the given image, you should notice that the base of the triangle is also the diameter of the circle. In addition, the height of the triangle is also the radius of the circle.

Thus, we can find the area of the triangle.

$\text{Area of Triangle}=\frac{1}{2}(\text{base}\times\text{height})$

$\text{Area of Triangle}=\frac{1}{2}(28 \times 14)=196$

Next, recall how to find the area of a circle.

$\text{Area of Circle}=\pi\times\text{radius}^2$

$\text{Area of Circle}=\pi\times 14^2=196 \pi$

To find the area of the shaded region, subtract the two areas.

$\text{Area of Shaded Region}=\text{Area of Circle}-\text{Area of Triangle}$

$\text{Area of Shaded Region}=196\pi-196=419.75$

Make sure to round to places after the decimal.

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Example Question #1 : How To Find The Area Of An Acute / Obtuse Isosceles Triangle

An isosceles triangle is placed in a circle as shown by the figure below.

If the diameter of the circle is , find the area of the shaded region.

Possible Answers:

173.47

198.52

189.62

174.55

Correct answer:

173.47

Explanation:

From the given image, you should notice that the base of the triangle is also the diameter of the circle. In addition, the height of the triangle is also the radius of the circle.

Thus, we can find the area of the triangle.

$\text{Area of Triangle}=\frac{1}{2}(\text{base}\times\text{height})$

$\text{Area of Triangle}=\frac{1}{2}(18 \times 9)=81$

Next, recall how to find the area of a circle.

$\text{Area of Circle}=\pi\times\text{radius}^2$

$\text{Area of Circle}=\pi\times 9^2= 81\pi$

To find the area of the shaded region, subtract the two areas.

$\text{Area of Shaded Region}=\text{Area of Circle}-\text{Area of Triangle}$

$\text{Area of Shaded Region}=81\pi-81=173.47$

Make sure to round to places after the decimal.

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Example Question #21 : Triangles

An isosceles triangle is placed in a circle as shown by the figure below.

If the diameter of the circle is , find the area of the shaded region.

Possible Answers:

401.25

372.90

355.21

361.93

Correct answer:

361.93

Explanation:

From the given image, you should notice that the base of the triangle is also the diameter of the circle. In addition, the height of the triangle is also the radius of the circle.

Thus, we can find the area of the triangle.

$\text{Area of Triangle}=\frac{1}{2}(\text{base}\times\text{height})$

$\text{Area of Triangle}=\frac{1}{2}(26 \times 13)=169$

Next, recall how to find the area of a circle.

$\text{Area of Circle}=\pi\times\text{radius}^2$

$\text{Area of Circle}=\pi\times 13^2= 169\pi$

To find the area of the shaded region, subtract the two areas.

$\text{Area of Shaded Region}=\text{Area of Circle}-\text{Area of Triangle}$

$\text{Area of Shaded Region}=169\pi-169=361.93$

Make sure to round to places after the decimal.

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Example Question #1 : How To Find The Area Of An Acute / Obtuse Isosceles Triangle

An isosceles triangle is placed in a circle as shown by the figure below.

If the diameter of the circle is , find the area of the shaded region.

Possible Answers:

1036.53

1121.77

980.27

1092.14

Correct answer:

1036.53

Explanation:

From the given image, you should notice that the base of the triangle is also the diameter of the circle. In addition, the height of the triangle is also the radius of the circle.

Thus, we can find the area of the triangle.

$\text{Area of Triangle}=\frac{1}{2}(\text{base}\times\text{height})$

$\text{Area of Triangle}=\frac{1}{2}(44 \times 22)=484$

Next, recall how to find the area of a circle.

$\text{Area of Circle}=\pi\times\text{radius}^2$

$\text{Area of Circle}=\pi\times 22^2= 484\pi$

To find the area of the shaded region, subtract the two areas.

$\text{Area of Shaded Region}=\text{Area of Circle}-\text{Area of Triangle}$

$\text{Area of Shaded Region}=484\pi-484=1036.53$

Make sure to round to places after the decimal.

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Example Question #21 : Triangles

An isosceles triangle is placed in a circle as shown by the figure below.

If the diameter of the circle is , find the area of the shaded region.

Possible Answers:

1145.52

1190.02

1098.98

1132.90

Correct answer:

1132.90

Explanation:

From the given image, you should notice that the base of the triangle is also the diameter of the circle. In addition, the height of the triangle is also the radius of the circle.

Thus, we can find the area of the triangle.

$\text{Area of Triangle}=\frac{1}{2}(\text{base}\times\text{height})$

$\text{Area of Triangle}=\frac{1}{2}(46 \times 23)=529$

Next, recall how to find the area of a circle.

$\text{Area of Circle}=\pi\times\text{radius}^2$

$\text{Area of Circle}=\pi\times 23^2= 529\pi$

To find the area of the shaded region, subtract the two areas.

$\text{Area of Shaded Region}=\text{Area of Circle}-\text{Area of Triangle}$

$\text{Area of Shaded Region}=529\pi-529=1132.90$

Make sure to round to places after the decimal.

Report an Error

Example Question #22 : Triangles

An isosceles triangle is placed in a circle as shown by the figure below.

If the diameter of the circle is , find the area of the shaded region.

Possible Answers:

498.20

402.08

481.86

441.12

Correct answer:

481.86

Explanation:

From the given image, you should notice that the base of the triangle is also the diameter of the circle. In addition, the height of the triangle is also the radius of the circle.

Thus, we can find the area of the triangle.

$\text{Area of Triangle}=\frac{1}{2}(\text{base}\times\text{height})$

$\text{Area of Triangle}=\frac{1}{2}(30 \times 15)=225$

Next, recall how to find the area of a circle.

$\text{Area of Circle}=\pi\times\text{radius}^2$

$\text{Area of Circle}=\pi\times 15^2= 225\pi$

To find the area of the shaded region, subtract the two areas.

$\text{Area of Shaded Region}=\text{Area of Circle}-\text{Area of Triangle}$

$\text{Area of Shaded Region}=225\pi-225=481.86$

Make sure to round to places after the decimal.

Report an Error

Example Question #23 : Triangles

An isosceles triangle is placed in a circle as shown by the figure below.

If the diameter of the circle is , find the area of the shaded region.

Possible Answers:

1190.21

1181.21

1447.72

1098.21

Correct answer:

1447.72

Explanation:

From the given image, you should notice that the base of the triangle is also the diameter of the circle. In addition, the height of the triangle is also the radius of the circle.

Thus, we can find the area of the triangle.

$\text{Area of Triangle}=\frac{1}{2}(\text{base}\times\text{height})$

$\text{Area of Triangle}=\frac{1}{2}(52 \times 26)=676$

Next, recall how to find the area of a circle.

$\text{Area of Circle}=\pi\times\text{radius}^2$

$\text{Area of Circle}=\pi\times 26^2= 676\pi$

To find the area of the shaded region, subtract the two areas.

$\text{Area of Shaded Region}=\text{Area of Circle}-\text{Area of Triangle}$

$\text{Area of Shaded Region}=676\pi-676=1447.72$

Make sure to round to places after the decimal.

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Example Question #461 : Intermediate Geometry

A triangle is placed in a parallelogram so that they share a base.

If the height of the triangle is half the height of the parallelogram, find the area of the shaded region.

Possible Answers:

1.25

0.75

1.5

Correct answer:

1.5

Explanation:

In order to find the area of the shaded region, we will need to find the areas of the triangle and of the parallelogram.

First, recall how to find the area of a parallelogram.

$\text{Area of Parallelogram}=\text{base}\times\text{height}$

$\text{Area of Parallelogram}=2\times 1 = 2$

Next, recall how to find the area of a triangle.

$\text{Area of Triangle}=\frac{1}{2}(\text{base}\times\text{height})$

Now, find the height of the triangle.

$\text{Height of Triangle}=\frac{\text{Height of Parallelogram}}{2}$

$\text{Height of Triangle}=\frac{1}{2}$

Plug this value in to find the area of the triangle.

$\text{Area of Triangle}=\frac{1}{2}(2\times \frac{1}{2})=\frac{1}{2}=0.5$

Subtract the two areas to find the area of the shaded region.

$\text{Area of Shaded Region}=\text{Area of Parallelogram}-\text{Area of Triangle}$

$\text{Area of Shaded Region}=2-0.5=1.5$

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Example Question #25 : Triangles

Parallel 2

Refer to the above diagram. m ||n .

True or false: From the information given, it follows that $\bigtriangleup AVB \sim \bigtriangleup CVD$ .

Possible Answers:

False

True

Correct answer:

True

Explanation:

By the Angle-Angle Similarity Postulate, if two pairs of corresponding angles of a triangle are congruent, the triangles themselves are similar.

$\angle AVB$ and $\angle CVD$ are a pair of vertical angles, having the same vertex and having sides opposite each other. As such, $\angle AVB \cong \angle CVD$ .

$\angle ABV$ and $\angle CDV$ are alternating interior angles formed by two parallel lines and cut by a transversal . As a consequence, $\angle ABV \cong \angle CDV$ .

The conditions of the Angle-Angle Similarity Postulate are satisfied, and it holds that $\bigtriangleup AVB \sim \bigtriangleup CVD$ .

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All Intermediate Geometry Resources

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Intermediate Geometry : Intermediate Geometry

Study concepts, example questions & explanations for Intermediate Geometry

All Intermediate Geometry Resources

Example Questions

Example Question #6 : How To Find The Area Of An Acute / Obtuse Isosceles Triangle

Example Question #7 : How To Find The Area Of An Acute / Obtuse Isosceles Triangle

Example Question #1 : How To Find The Area Of An Acute / Obtuse Isosceles Triangle

Example Question #21 : Triangles

Example Question #1 : How To Find The Area Of An Acute / Obtuse Isosceles Triangle

Example Question #21 : Triangles

Example Question #22 : Triangles

Example Question #23 : Triangles

Example Question #461 : Intermediate Geometry

Example Question #25 : Triangles

All Intermediate Geometry Resources

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